Show your work for the following questions- 11- Consider the following.docx

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1) Using the given enthalpy values for two reactions, the heat of reaction (ΔH) for CO2(g) → CO(g) + 1/2O2(g) is calculated to be 283 kJ. 2) For the combustion of 1.00 g of glucose, the enthalpy of combustion is calculated to be -15.52 kJ/g and the molar enthalpy of combustion is calculated to be -2796.4 kJ/mol. 3) For the combustion of 0.1375 g of Mg, using the temperature change and heat capacity, the molar enthalpy of combustion of Mg is calculated.

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Show your work for the following questions. 11. Consider the following two reactions and
enthalpies: 1) 2c(s) + 0:(g)--2Co() 2) C(s) +(gCO2) AH220 kJ/mol AH393 kJ/mol From these
two reactions, calculate the heat effeet (enthalpy ehange AH) for the reaction: 3) CO2(g)-
CO(g)+HO(g) 12. When 1.00 g glucose (CH 0 180 g/mo is combusted in an open calorimeter
with calorimeter constant 8000 J/K, the temperature rises from 19.20 Â°C to 21.14 C^Calculate:
a) The enthalpy of combustion of AHseabuionsglucose) in kJ/g b) The molar enthalpy of
combustion of glucose 13. (Chang 6.37) A 0.1375 g sample of Mg(s) is burned (reacted with O2)
in a calorimeter. The heat capacity C of the calorimeter + content is 3024 JrCThe temperature
increases by 1.126 *C. Calculate the Molar enthalpy of combustion, AH of Mg (kJ/moD. This
problem is worded slightly different from the version in Chang, because in a "bomb calorimeter
we would measure ??, not ??!) Hint: We are looking for Akl in kliua, i.e., the enthalpy change
for the combustion of 1 malofMg. 14. 25.0 mL.0.500 M .HCI(ag) is added to 75.0 mL 0.500 M
NaO in an open adiabatic styrofoam "coffee cup calorimeter". The temperature rises by1.68
JC.Calculate the molar enthalpy of neutralization on HCl The heat capacity of the Styrofoam cup
can be neglected. The heat capacity of 1 mL solution can be assumed to be equal to the heat
capacity of 1 mL (1 g) water.
Solution
11)
from the given data
CO2(g) ----> c(s) + o2(g)Â Â Â DH1 = +393 kj/mol
(2C(s) + O2(g) ---> 2CO(g)Â Â Â DH2 = -220 kj/mol)*1/2
---------------------------------------------------
CO2(g) ----> CO(g) + 1/2O2(g) DH = DH1+DH2*1/2
----------------------------------------------------
DHrxn = 393-220*1/2 = 283 kj
12) energy released (q) = C*DT
= 8000*(21.14-19.2)

= 15520 j
= 15.52 kj
a) DHcombustion = -q/w
= -15.52/1
= -15.52 kj/g
b) No of mol of glucose = w/Mwt = 1/180 = 0.00555 mol
molar enthalpy of combustion = -q/n
= -15.52/0.00555
= -2796.4 kj/mol

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